A301930 G.f. A(x,y) satisfies: A(x,y) = x * (1 + y*A(x,y)*A'(x,y)) / (1 + A(x,y)*A'(x,y)), where A'(x,y) = d/dx A(x,y).
1, -1, 1, 4, -7, 3, -25, 63, -52, 14, 200, -661, 808, -432, 85, -1890, 7754, -12586, 10090, -3989, 621, 20248, -99450, 201726, -216125, 128869, -40504, 5236, -240069, 1375831, -3354625, 4508559, -3604985, 1713731, -448122, 49680, 3102000, -20349633, 58049510, -94012374, 94504280, -60352776, 23900178, -5362906, 521721, -43226590, 319817454, -1046234664, 1985688420, -2408884136, 1936407600, -1031098592, 350561508, -69025155, 5994155
Offset: 1
Examples
G.f.: A(x,y) = Sum_{n>=1} Sum_{k=0..n-1} T(n,k)*x^n*y^k = x + (-1 + y)*x^2 + (4 - 7*y + 3*y^2)*x^3 + (-25 + 63*y - 52*y^2 + 14*y^3)*x^4 + (200 - 661*y + 808*y^2 - 432*y^3 + 85*y^4)*x^5 + (-1890 + 7754*y - 12586*y^2 + 10090*y^3 - 3989*y^4 + 621*y^5)*x^6 + ...
such that A = A(x,y) satisfies A = x*(1 + y*A*A')/(1 + A*A').
This triangle of coefficients T(n,k) in A(x,y) begins:
[1];
[-1, 1];
[4, -7, 3];
[-25, 63, -52, 14];
[200, -661, 808, -432, 85];
[-1890, 7754, -12586, 10090, -3989, 621];
[20248, -99450, 201726, -216125, 128869, -40504, 5236];
[-240069, 1375831, -3354625, 4508559, -3604985, 1713731, -448122, 49680];
[3102000, -20349633, 58049510, -94012374, 94504280, -60352776, 23900178, -5362906, 521721];
[-43226590, 319817454, -1046234664, 1985688420, -2408884136, 1936407600, -1031098592, 350561508, -69025155, 5994155]; ...
SPECIAL CASES.
G.f. C(x) of column 0 satisfies: C = x - C'*C^2, and begins C(x) = x - x^2 + 4*x^3 - 25*x^4 + 200*x^5 - 1890*x^6 +...
G.f. D(x) of the main diagonal satisfies: D = x + x*D'*D, and begins D(x) = x + x^2 + 3*x^3 + 14*x^4 + 85*x^5 + 621*x^6 + ...
At y = 2, the row polynomials evaluate to form the Catalan numbers:
1 = 1;
1 = -1 + 1*2;
2 = 4 + -7*2 + 3*2^2;
5 = -25 + 63*2 + -52*2^2 + 14*2^3;
14 = 200 + -661*2 + 808*2^2 + -432*2^3 + 85*2^4;
42 = -1890 + 7754*2 + -12586*2^2 + 10090*2^3 + -3989*2^4 + 621*2^5; ...
illustrating: C(2*n-1,n-1)/(2*n-1) = Sum_{k=0..n-1} T(n,k) * 2^k.
Note: when the g.f. A(x,y) is evaluated at y < 2 and y not= 1, the resulting power series in x will have negative coefficients somewhere in the expansion.
Links
Programs
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PARI
{T(n,k) = my(A=x); for(i=1,n, A = x*(1 + y*A*A')/(1 + A*A' +x*O(x^n))); polcoeff(polcoeff(A,n,x),k,y)} /* Print as a triangle */ for(n=1,10,for(k=0,n-1, print1(T(n,k),", "));print("")) /* Print as a flattened triangle: */ for(n=1,10, for(k=0,n-1, print1(T(n,k),", "); );)
Formula
Column 0 equals A088716 (signed).
Main diagonal equals A182304.
Row sums are zeros after the initial row.
Absolute row sums = A301931.
Sum_{k=0..n-1} T(n,k) * 2^k = C(2*n-1,n-1)/(2*n-1) = A000108(n-1) for n>=1.
Sum_{k=0..n-1} T(n,k) * 3^k = A301932(n) for n>=1.
Sum_{k=0..n-1} T(n,k) * 4^k = A301933(n) for n>=1.
Limit of largest real root of row polynomials converges to 2.
Comments